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“在4.6.2节中我们已经知道,如果一个变换中仅存在平移和旋转变换,那么这个变换的逆矩阵就等于它的转置矩阵”,这里“平移”是不是要换成统一缩放或删除呢?TBN矩阵之所以是正交的体现在,由于对坐标轴的平移变换没有作用,因此只需要3*3的旋转矩阵,缩放也不需要,这里和第四章叙述的内容似乎并不一致。
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“在4.6.2节中我们已经知道,如果一个变换中仅存在平移和旋转变换,那么这个变换的逆矩阵就等于它的转置矩阵”,这里“平移”是不是要换成统一缩放或删除呢?TBN矩阵之所以是正交的体现在,由于对坐标轴的平移变换没有作用,因此只需要3*3的旋转矩阵,缩放也不需要,这里和第四章叙述的内容似乎并不一致。
The text was updated successfully, but these errors were encountered: